How Can I improve the result of Surrogateopt Optimization?

7 Ansichten (letzte 30 Tage)
Mehmet Türker TAKCI
Mehmet Türker TAKCI am 13 Feb. 2022
Hi all, I have a mixed integer nonlinear problem with a lot of constraints and used surrogateopt to find the minimum cost function. It gives a result but It doesn't acceptable one. I changed some default parameters like MaxFunctionEvaluations, MinSampleDistance or MinSurrogatePoints. At the same time, I have added InitialPoints but the result didn't change. I solved the same problem with fmincon without using integer constraints and found a good solution. I modified this result as initialpoints for the surrogate optimization but It still gives the same bad result. It didn't consider the initial points. When I call the function with intialpoints, It gives a better result. For example;
opts = optimoptions('surrogateopt','PlotFcn','surrogateoptplot','InitialPoints',Xint,'MaxFunctionEvaluations',6000);
rng default
[x1,fval,exitflag,output] = surrogateopt(f,LB,UB,intcon,A,b,Aeq,beq,opts);
fval= 1000 $
fval_new=f(Xint) = 800$
So, why can't the surrogate find a good result at least as good as the intial points I gave? How can I improve my result?

Antworten (2)

Alan Weiss
Alan Weiss am 14 Feb. 2022
Is your initial point Xint feasible? In other words, does it satisfy
  • Xint >= lb
  • Xint <= ub
  • Aeq*Xint' = beq
  • A*Xint' <= b
  • f.Ineq(Xint) <= 0
If Xint satisfies all of those constraints then I would be surprised that surrogateopt returns a worse solution thant Xint.
Alan Weiss
MATLAB mathematical toolbox documentation
  3 Kommentare
Alan Weiss
Alan Weiss am 16 Feb. 2022
I would appreciate seeing your code. I cannot understand how surrogateopt would return a worse solution than a feasible initial point.
Alan Weiss
MATLAB mathematical toolbox documentation
Mehmet Türker TAKCI
Mehmet Türker TAKCI am 18 Feb. 2022
Hi Alan, did you have a chance to examine the code?

Melden Sie sich an, um zu kommentieren.


Mehmet Türker TAKCI
Mehmet Türker TAKCI am 16 Feb. 2022
Bearbeitet: Mehmet Türker TAKCI am 2 Mär. 2022
The mathematical equations are :
  2 Kommentare
Alan Weiss
Alan Weiss am 18 Feb. 2022
I believe that the issue arises because your initial point Xint is not feasible, as you will see if you execute
max(abs(AeqXint - beq))
You get a value of 0.0027, which is large enough for the solver to deem the point infeasible. To examine the issue further, execute
[rr ss] = max(abs(AeqXint - beq)) % ss is the index where it occurs
AeqXint(ss)
beq(ss)
Alan Weiss
MATLAB mathematical toolbox documentation
Mehmet Türker TAKCI
Mehmet Türker TAKCI am 19 Feb. 2022
Hi Alan, thanks for your help.

Melden Sie sich an, um zu kommentieren.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by